3.1.35 \(\int \frac {a+b \cosh ^{-1}(c x)}{x^3 (d-c^2 d x^2)} \, dx\) [35]

Optimal. Leaf size=118 \[ \frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+\frac {2 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b c^2 \text {PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b c^2 \text {PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d} \]

[Out]

1/2*(-a-b*arccosh(c*x))/d/x^2+2*c^2*(a+b*arccosh(c*x))*arctanh((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d+1/2*b*c^
2*polylog(2,-(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d-1/2*b*c^2*polylog(2,(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d
+1/2*b*c*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d/x

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Rubi [A]
time = 0.14, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5932, 5916, 5569, 4267, 2317, 2438, 97} \begin {gather*} \frac {2 c^2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+\frac {b c^2 \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b c^2 \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d}+\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/(x^3*(d - c^2*d*x^2)),x]

[Out]

(b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(2*d*x) - (a + b*ArcCosh[c*x])/(2*d*x^2) + (2*c^2*(a + b*ArcCosh[c*x])*ArcT
anh[E^(2*ArcCosh[c*x])])/d + (b*c^2*PolyLog[2, -E^(2*ArcCosh[c*x])])/(2*d) - (b*c^2*PolyLog[2, E^(2*ArcCosh[c*
x])])/(2*d)

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5916

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[-d^(-1), Subst[I
nt[(a + b*x)^n/(Cosh[x]*Sinh[x]), x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] &
& IGtQ[n, 0]

Rule 5932

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcCosh[c*x])^n/(d*f*(m + 1))), x] + (Dist[c^2*((m + 2*p + 3)/(f^2*(
m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcCosh[c*x])^n, x], x] + Dist[b*c*(n/(f*(m + 1)))*Simp[(d +
e*x^2)^p/((1 + c*x)^p*(-1 + c*x)^p)], Int[(f*x)^(m + 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(a + b*ArcCos
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \cosh ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )} \, dx &=-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+c^2 \int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx+\frac {(b c) \int \frac {1}{x^2 \sqrt {-1+c x} \sqrt {1+c x}} \, dx}{2 d}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}-\frac {c^2 \text {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}-\frac {\left (2 c^2\right ) \text {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+\frac {2 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {\left (b c^2\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d}-\frac {\left (b c^2\right ) \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+\frac {2 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {\left (b c^2\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {\left (b c^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}\\ &=\frac {b c \sqrt {-1+c x} \sqrt {1+c x}}{2 d x}-\frac {a+b \cosh ^{-1}(c x)}{2 d x^2}+\frac {2 c^2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b c^2 \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b c^2 \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 144, normalized size = 1.22 \begin {gather*} -\frac {\frac {a}{x^2}-2 a c^2 \log (x)+a c^2 \log \left (1-c^2 x^2\right )+b c^2 \left (-\frac {\sqrt {\frac {-1+c x}{1+c x}} (1+c x)}{c x}+\frac {\cosh ^{-1}(c x)}{c^2 x^2}+2 \cosh ^{-1}(c x) \log \left (1-e^{-2 \cosh ^{-1}(c x)}\right )-2 \cosh ^{-1}(c x) \log \left (1+e^{-2 \cosh ^{-1}(c x)}\right )+\text {PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-2 \cosh ^{-1}(c x)}\right )\right )}{2 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c*x])/(x^3*(d - c^2*d*x^2)),x]

[Out]

-1/2*(a/x^2 - 2*a*c^2*Log[x] + a*c^2*Log[1 - c^2*x^2] + b*c^2*(-((Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x))/(c*x))
+ ArcCosh[c*x]/(c^2*x^2) + 2*ArcCosh[c*x]*Log[1 - E^(-2*ArcCosh[c*x])] - 2*ArcCosh[c*x]*Log[1 + E^(-2*ArcCosh
[c*x])] + PolyLog[2, -E^(-2*ArcCosh[c*x])] - PolyLog[2, E^(-2*ArcCosh[c*x])]))/d

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Maple [A]
time = 4.49, size = 283, normalized size = 2.40

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a}{2 d \,c^{2} x^{2}}+\frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {b \sqrt {c x +1}\, \sqrt {c x -1}}{2 d c x}-\frac {b}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right )}{2 d \,c^{2} x^{2}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}+\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{d}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}\right )\)	\(283\)
default	\(c^{2} \left (-\frac {a}{2 d \,c^{2} x^{2}}+\frac {a \ln \left (c x \right )}{d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {b \sqrt {c x +1}\, \sqrt {c x -1}}{2 d c x}-\frac {b}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right )}{2 d \,c^{2} x^{2}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}+\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{d}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}\right )\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(c*x))/x^3/(-c^2*d*x^2+d),x,method=_RETURNVERBOSE)

[Out]

c^2*(-1/2*a/d/c^2/x^2+a/d*ln(c*x)-1/2*a/d*ln(c*x-1)-1/2*a/d*ln(c*x+1)+1/2*b/d/c/x*(c*x+1)^(1/2)*(c*x-1)^(1/2)-
1/2*b/d-1/2*b/d*arccosh(c*x)/c^2/x^2-b/d*arccosh(c*x)*ln(1+c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))-b/d*polylog(2,-c*x
-(c*x-1)^(1/2)*(c*x+1)^(1/2))+b/d*arccosh(c*x)*ln(1+(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)+1/2*b*polylog(2,-(c*x
+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/d-b/d*arccosh(c*x)*ln(1-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))-b/d*polylog(2,c*x+(c
*x-1)^(1/2)*(c*x+1)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^3/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/2*(c^2*log(c*x + 1)/d + c^2*log(c*x - 1)/d - 2*c^2*log(x)/d + 1/(d*x^2))*a - b*integrate(log(c*x + sqrt(c*x
+ 1)*sqrt(c*x - 1))/(c^2*d*x^5 - d*x^3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^3/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*arccosh(c*x) + a)/(c^2*d*x^5 - d*x^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a}{c^{2} x^{5} - x^{3}}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(c*x))/x**3/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a/(c**2*x**5 - x**3), x) + Integral(b*acosh(c*x)/(c**2*x**5 - x**3), x))/d

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x^3/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate(-(b*arccosh(c*x) + a)/((c^2*d*x^2 - d)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{x^3\,\left (d-c^2\,d\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(x^3*(d - c^2*d*x^2)),x)

[Out]

int((a + b*acosh(c*x))/(x^3*(d - c^2*d*x^2)), x)

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